MCQ
If $(1 - i)^n = 2^n,$ then $n = $
- A$1$
- ✓$0$
- C$- 1$
- DNone of these
$ \Rightarrow $ $|1 - i|^n = \,|2|^n$,$(\because \,\,2^n > 0)$
==> $\left[ \sqrt {{1^2} + {{( - 1)}^2}} \right]^n = 2^n$
==> $(\sqrt 2 )^n = 2^n$
==> $2^{n/2} = 2^n$
==> $\frac{n}{2} = n$
==>$n = 0$
Trick : By inspection, ${(1 - i)^0} = {2^0}\,\,\,\, \Rightarrow 1 = 1$.
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